VECTOR MECH...,STAT.+DYNA.(LL)-W/ACCESS
VECTOR MECH...,STAT.+DYNA.(LL)-W/ACCESS
11th Edition
ISBN: 9781259633133
Author: BEER
Publisher: MCG
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Chapter 3.3, Problem 3.71P

(a)

To determine

The moment of the couple (M) formed by two forces by resolving each force into horizontal and vertical components and adding the moments of the two resulting couples.

(a)

Expert Solution
Check Mark

Answer to Problem 3.71P

The moment of the couple (M) formed by two forces by resolving each force into horizontal and vertical components and adding the moments of the two resulting couples is 6.19Nm(clockwise)_.

Explanation of Solution

Given information:

The applied force at point B (FB) is 40 N.

The applied force at point C (FC) is 40 N.

The length of AB (x) is 390 mm.

The length of BC (y) is 270 mm.

The angle of the inclined lever (θ) is 55°.

The angle of the force acting at point C (α) is 20°.

Calculation:

Draw the free body diagram of the lever as in Figure (1).

VECTOR MECH...,STAT.+DYNA.(LL)-W/ACCESS, Chapter 3.3, Problem 3.71P

Calculate the vertical height of BC (d1) by resolving in vertical direction using the relation:

d1=ysinθ

Substitute 270 mm for y and 55° for θ.

d1=(270mm×11000mmm)sin55°=0.22117m

Calculate the horizontal height of BC (d2) by resolving in horizontal direction using the relation:

d2=ycosθ

Substitute 270 mm for y and 55° for θ.

d2=(270mm×11000mmm)cos55°=0.154866m

Calculate the horizontal reaction at C (Cx) by resolving in horizontal direction using the relation:

Cx=FC×cosα

Substitute 40 N for FC and 20° for α.

Cx=40×cos20=37.588N

Calculate the vertical reaction at C (Cy) by resolving in vertical direction using the relation:

Cy=FC×sinα

Substitute 40 N for FC and 20° for α.

Cy=40×sin20=13.6808N

Find the moment of the couple (M):

Take the moment about B.

M=d1Cx+d2Cy

Substitute 0.22117 m for d1, 0.154866 m for d2, 37.588 N for Cx, and 13.6808 N for Cy

M=(0.22117×37.588)+(0.154866×13.6808)=6.19Nm=6.19Nm(clockwise)

Thus, the moment of the couple (M) formed by two forces by resolving each force into horizontal and vertical components and adding the moments of the two resulting couples is 6.19Nm(clockwise)_.

(b)

To determine

The moment of the couple (M) formed by two forces by using the perpendicular distance between the two forces.

(b)

Expert Solution
Check Mark

Answer to Problem 3.71P

The moment of the couple (M) formed by two forces by using the perpendicular distance between the two forces is 6.1946Nm(k)(clockwise)_.

Explanation of Solution

Given information:

The applied force at point B (FB) is 40 N.

The applied force at point C (FC) is 40 N.

The length of AB (x) is 390 mm.

The length of BC (y) is 270 mm.

The angle of the inclined lever (θ) is 55°.

The angle of the force acting at point C (α) is 20°.

Calculation:

Calculate the distance (d) between the two forces using the relation:

d=ysin(θα)

Substitute 270 mm for y, 55° for θ and 20° for α.

d=(270mm×11000mmm)sin(55°20°)=0.154866m

Calculate the moment of the couple (M) formed by two forces by using the perpendicular distance between the two forces using the relation:

M=Fd(k)

Substitute 40 N for F and 0.154866 m for d.

M=40×0.154866(k)=6.1946Nm(k)=6.1946Nm(k)(clockwise)

Thus, the moment of the couple (M) formed by two forces by using the perpendicular distance between the two forces is 6.1946Nm(k)(clockwise)_.

(c)

To determine

The moment of the couple (M) formed by summing the moments of two forces about point A.

(c)

Expert Solution
Check Mark

Answer to Problem 3.71P

The moment of the couple (M) formed by summing the moments of two forces about point A is (6.195Nm)k(clockwise)_.

Explanation of Solution

Given information:

The applied force at point B (FB) is 40 N.

The applied force at point C (FC) is 40 N.

The length of AB (x) is 390 mm.

The length of BC (y) is 270 mm.

The angle of the inclined lever (θ) is 55°.

The angle of the force acting at point C (α) is 20°.

Calculation:

Calculate the position vector of from point B to point A (rB/A) using the relation:

rB/A=xcosθi+xsinθj+0k=xcosθi+xsinθj

Substitute 390 mm for x and 55° for θ.

rB/A=(390mm×11000mmm)cos55°i+(390mm×11000mmm)sin55°j=0.39cos55°i+0.39sin55°j

Calculate the force at B by resolving in horizontal and vertical direction using the relation:

FB=FBcosαiFBsinαj+0

Substitute 40 N for FB and 20° for α.

FB=40cos20°i40sin20°j+0=40cos20°i40sin20°j

Calculate the position vector of from point C to point A (rC/A) using the relation:

rC/A=(x+y)cosθi+(x+y)sinθj+0k=(x+y)cosθi+(x+y)sinθj

Substitute 390 mm for x, 270 mm for y and 55° for θ.

rC/A={[(390mm×11000mmm)+(270mm×11000mmm)]cos55°}i+{[(390mm×11000mmm)+(270mm×11000mmm)]sin55°}j=0.66cos55°i+0.66sin55°j

Calculate the force at C by resolving in horizontal and vertical direction using the relation:

FC=FCcosαi+FCsinαj+0

Substitute 40 N for FC and 20° for α.

FC=40cos20°i+40sin20°j+0=40cos20°i+40sin20°j

Calculate the moment of the couple (M) formed by summing the moments of two forces about point A using the relation:

Take the moment about A.

MA=rA×FM=(rB/A×FB)+(rC/A×FC)

Substitute 0.39cos55°i+0.39sin55°j for rB/A, 0.66cos55°i+0.66sin55°j for rC/A, 40cos20°i40sin20°j for FB, and 40cos20°i+40sin20°j for FC.

M=(0.39m)(40N)|ijkcos55°sin55°0cos20°sin20°0|+(0.66m)(40N)|ijkcos55°sin55°0cos20°sin20°0|={(0.39m)(40N)[(0)(0)(0.1962+0.7698)]+(0.66m)(40N)[00(0.19620.7698)]}=(8.94815.143)k=(6.195Nm)k=(6.195Nm)k(clockwise)

Thus, the moment of the couple (M) formed by summing the moments of two forces about point A is (6.195Nm)k(clockwise)_.

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Chapter 3 Solutions

VECTOR MECH...,STAT.+DYNA.(LL)-W/ACCESS

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